Posts Tagged ‘force times distance’

We’ve been discussing the mechanical advantage that compound pulleys provide to humans during lifting operations and last time we hit upon the fact that there comes a point of diminished return, a reality that engineers must negotiate in their mechanical designs. Today we’ll discuss one of the undesirable tradeoffs that results in a diminished return within a compound pulleyarrangement when we compute the length of rope the Grecian man we’ve been following must grapple in order to lift his urn. What we’ll discover is a situation of mechanical overkill – like using a steamroller to squash a bug.

As presented in a past blog, the equations for work input, WI, and work output, WO, we’ll be using are,

WI = F × d2

WO = W × d1

Now, ideally, in a compound pulley no friction exists in the wheels to impede the rope’s movement, and that will be our scenario today. Our next blog will deal with the more complex situation where friction is present. So for our example today, with no friction present, work input equals output…

WI = WO

… and this fact allows us to develop an equation in terms of the rope length/distance factors in our compound pulley assembly, represented by d1 and d2, …

F × d2 = W × d1

d2 ÷ d1 = W ÷ F

Now, from our last blog we know that W divided by F represents the mechanical advantage, MA, to Mr. Toga of using the compound pulley, which was found to be 16, equivalent to the sections of rope directly supporting the urn. We’ll set the distance factors up in relation to MA, and the equation becomes…

d2÷ d1 = MA

d2 = MA × d1

d2 = 16 ×2 feet = 32 feet

What we discover is that in order to raise the urn 2 feet, our Grecian friend must manipulate 32 feet of rope – which would only make sense if he were lifting something far heavier than a 40 pound urn.

In reality, WI does not equal WO, due to the inevitable presence of friction. Next time we’ll see how friction affects the mechanical advantage in our compound pulley.